From C6 Rate & Equilibrium you already know reversible reactions, that equilibrium needs a closed system, and Le Chatelier’s principle in words. A-Level makes it exact: dynamic equilibrium gets a precise definition (equal rates), Le Chatelier predictions become two-step mark-scheme answers, and “yield” becomes a number you calculate — Kc from concentrations, and (for A-level) Kp from partial pressures. Hold one distinction the whole way through: some changes move the position of equilibrium; only temperature changes the value of the constant.
Dynamic equilibrium
Start a reversible reaction in a closed system and two things happen at once: the forward reaction slows down as the reactants are used up, while the reverse reaction speeds up as products build. Sooner or later the two rates become equal — that is equilibrium. From the outside nothing seems to change, but neither reaction has stopped: reactants are still turning into products and products back into reactants, just as fast as each other. That constant two-way traffic is what “dynamic” means.
Dynamic equilibrium — in a closed system, the forward and reverse reactions occur at the same rate, so the concentrations of reactants and products remain constant.
That last point is the one examiners see missed most: equilibrium means the concentrations stay constant, not that they are equal. Reactants and products usually sit at very different concentrations — they just stop changing.
Equilibrium can only be reached in a closed system — one where nothing enters or leaves. The moment a product can escape, the story changes:
- Equal rates, not equal concentrations. Examiner reports flag this repeatedly — dynamic equilibrium is defined by the forward and reverse reactions going at the same rate. “The concentrations are equal” is wrong.
- Closed system. Nothing can enter or leave, or equilibrium can never be reached.
- Still reacting. “Dynamic” means both reactions continue — the reaction has not stopped, it has balanced.
GCSE groundwork: C6 — reversible reactions & equilibrium.
🧪 Exam-style questions
Which statement is not always correct for a reaction at equilibrium? reactants ⇌ products Tick (✓) one box.
Give two features of a reaction in dynamic equilibrium.
Show answer
The forward and reverse reactions proceed at equal rates. 1 mark
The concentrations of reactants and products remain constant. 1 mark
Do not accept “the concentrations are equal” or “the concentrations are the same” — the mark is for constant.
Source: AQA A-Level Chemistry past papers.
Le Chatelier’s principle
Le Chatelier’s principle is the tool for predicting which way an equilibrium moves when you disturb it.
Le Chatelier’s principle — when a change is made to a system at equilibrium, the position of equilibrium shifts to oppose the change.
“Shifts to oppose” gives you every prediction, once you read the equation. Take each lever in turn:
| Change | Which way the position shifts | Why |
|---|---|---|
| Increase a reactant’s concentration | towards the products | to use up the added reactant |
| Increase the pressure (gases) | to the side with fewer moles of gas | to reduce the pressure |
| Increase the temperature | in the endothermic direction | to absorb the added heat |
| Add a catalyst | no shift | it speeds both directions equally — position and yield unchanged |
All three panels below use N2(g) + 3H2(g) ⇌ 2NH3(g) — the forward reaction is exothermic, with 4 mol of gas on the left and 2 mol on the right.
“For an exothermic forward reaction, explain the effect of increasing the temperature on the equilibrium yield of product.” A full-mark answer is exactly two steps:
Step 1 — direction of shift: the position of equilibrium shifts in the endothermic (reverse) direction to oppose the increase in temperature. 1 mark
Step 2 — the consequence: so the equilibrium yield of product decreases. 1 mark
Every “explain the effect” answer follows this shape: name the direction the position moves, then state what happens to the quantity the question asks about (yield, amount, or value of K).
- Say “the position of equilibrium shifts”. Examiner reports flag answers that talk about the reaction “wanting to” oppose a change, or about the forward/reverse rates — write about the position moving.
- Two steps, every time. Direction of shift, then the consequence for yield / amount / value of K — a direction on its own does not score the second mark.
- A catalyst does nothing to the position. It speeds the forward and reverse reactions equally, so the yield is unchanged — equilibrium is just reached faster.
Interactive — the Le Chatelier sandbox: pull a lever, watch the position tip
GCSE groundwork: C6 — changing equilibrium conditions.
🧪 Exam-style questions
Methanol is made in this reaction, using a catalyst: CO(g) + 2H2(g) ⇌ CH3OH(g), ΔH = −91 kJ mol−1. Which change will decrease the equilibrium yield of methanol? Tick (✓) one box.
A gas-phase reaction is at equilibrium. When the pressure is increased, the yield of product decreases. State what can be deduced about the chemical equation for this equilibrium.
Show answer
There are more moles of gas on the product side than on the reactant side. 1 mark
Increasing the pressure shifts the position to the side with fewer moles of gas; since that lowered the yield, the products must be the side with more gas moles.
Tetrafluoroethene is made in this reaction: 2CHClF2(g) ⇌ C2F4(g) + 2HCl(g), ΔH = +128 kJ mol−1. State and explain the effect of using a higher temperature on the equilibrium yield of tetrafluoroethene.
Show answer
Effect: the yield of tetrafluoroethene increases. 1 mark
The position of equilibrium shifts in the endothermic (forward) direction 1 mark to oppose the increase in temperature / to lower the temperature. 1 mark
The forward reaction is endothermic (ΔH positive), so heating favours it. Note the two-step shape: direction of shift, then the reason it opposes the change.
Source: AQA A-Level Chemistry past papers.
Compromise conditions in industry
Le Chatelier tells a factory which conditions give the best yield — but the best yield is rarely the best economics. Real processes settle on a compromise between yield, rate and cost.
- Pressure. A high pressure boosts the yield when the products are the side with fewer moles of gas — but high-pressure plant is expensive to build and run, so a compromise pressure is used.
- Temperature. For an exothermic reaction a low temperature boosts the yield, but makes the reaction slow. A compromise temperature gives an economic rate at an acceptable yield.
- Catalyst. A catalyst does not change the yield, but it speeds the reaction so a lower temperature can be used — recovering some of the yield lost to the compromise, and cutting energy cost.
- Name the compromise, and give both reasons. A full answer states the yield reason and the rate/cost reason for each condition — not just “high pressure gives more product”.
- The catalyst does not raise the yield. Its economic value is letting you run at a lower temperature (faster, cheaper), not shifting the position.
Recap: a catalyst speeds the rate of attainment without moving the position, and whether the forward reaction gives out or takes in heat is set by its enthalpy change.
🧪 Exam-style questions
Hydrogen is prepared industrially using the reversible reaction between methane and steam: CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g), ΔH = +206 kJ mol−1. The reaction is carried out at 800 °C and a low pressure of 300 kPa with a nickel catalyst. Explain, in terms of equilibrium yield and cost, why these conditions are used.
Show answer
Temperature. The forward reaction is endothermic, so a high temperature shifts the position to the right and increases the yield; a very high temperature is costly in energy, so 800 °C is a compromise. up to 2 marks
Pressure. There are more moles of gas on the right (4) than the left (2), so a low pressure shifts the position to the right and increases the yield; a low pressure is also cheaper (less energy and less robust plant). up to 2 marks
Catalyst. The nickel catalyst speeds up the reaction (rate of attainment of equilibrium) without changing the yield, allowing a lower temperature to be used, which lowers the cost. up to 2 marks
Marked by levels — a top answer covers all three stages with a clear yield reason and a cost/rate reason for each.
Source: AQA A-Level Chemistry past papers.
Writing and calculating Kc
Le Chatelier gives you the direction; the equilibrium constant gives you the number. For a homogeneous equilibrium (everything in one phase), Kc is built from concentrations.
For a + b ⇌ c + d, the equilibrium constant Kc = [C]c[D]d ÷ [A]a[B]b, where [X] is the concentration of X in mol dm−3 at equilibrium, at constant temperature.
Products go on top, reactants on the bottom, each raised to its balancing number. The working routine is always the same three moves:
For N2(g) + 3H2(g) ⇌ 2NH3(g), a 2.00 dm3 vessel at equilibrium contains 0.20 mol N2, 0.60 mol H2 and 0.40 mol NH3. Find Kc.
Step 1 — convert amounts to concentrations (÷ volume):
[N2] = 0.10, [H2] = 0.30, [NH3] = 0.20 mol dm−3
Step 2 — substitute into the expression:
Kc = [NH3]2 ÷ ([N2][H2]3) = (0.20)2 ÷ (0.10 × (0.30)3)
Step 3 — evaluate, then derive the units:
Kc = 0.04 ÷ 0.0027 = 14.8 mol−2 dm6 (3 sf)
Units are built from the expression: (mol dm−3)2 ÷ (mol dm−3)4 = mol−2 dm6. They change with the stoichiometry — never assume “no units”.
If the equation has the same number of moles of gas on each side, every volume on the top of the Kc expression is matched by one on the bottom, so they cancel. You can then substitute the amounts in mol directly — no need to convert to concentrations — and Kc comes out with no units. For H2(g) + I2(g) ⇌ 2HI(g) (2 mol → 2 mol), Kc = [HI]2 ÷ ([H2][I2]) gives the same value from moles or concentrations. The ammonia example above does not balance (4 mol → 2 mol), so there you must divide by volume and the units survive.
- Divide by volume first. Examiner reports flag students substituting amounts (mol) instead of concentrations — convert with ÷ volume before the numbers go in (unless the moles of gas balance and the volumes cancel).
- Derive the units every time. Cancel the units of the expression for that exact stoichiometry — they are often not the same as the last question.
- Square brackets mean concentration in mol dm−3. Homogeneous means all species in one phase (all gas, or all aqueous).
Recap: converting moles to a concentration in mol dm−3 is the ÷ volume step from Amount of Substance.
Interactive — build the Kc expression, then find its value
Put each species on the correct line — products on top, reactants on the bottom.
-
Step 1 — convert each amount to a concentration (moles ÷ the volume), in mol dm−3:
-
Step 2 — substitute into Kc and evaluate (to 3 significant figures):
Kc = -
Step 3 — the units of Kc (cancel the units of the expression for this stoichiometry):
🧪 Exam-style questions
Methanol is made in this reaction: CO(g) + 2H2(g) ⇌ CH3OH(g). 2.0 mol of CO is mixed with 3.0 mol of H2 and reaches equilibrium containing 0.6 mol of methanol. What is the total amount, in mol, of gas at equilibrium? Tick (✓) one box.
When HF is added to water at 298 K: HF(aq) ⇌ H+(aq) + F−(aq). At equilibrium [HF] = 7.70 × 10−3 and [F−] = 2.30 × 10−3 mol dm−3. What is the value of Kc, in mol dm−3? Tick (✓) one box.
For A(aq) + 2B(aq) ⇌ C(aq), an equilibrium mixture has [A] = 0.48 mol dm−3, [C] = 0.62 mol dm−3 and Kc = 7.8 mol−2 dm6. Calculate [B] at equilibrium, to the appropriate number of significant figures.
Show answer
Kc = [C] ÷ ([A][B]2) → [B]2 = [C] ÷ (Kc[A])
Rearrange for [B]2 1 mark
[B]2 = 0.62 ÷ (7.8 × 0.48) = 0.1656
Evaluate [B] = √0.1656 = 0.4069 1 mark
[B] = 0.41 mol dm−3 (2 sf)
Answer to 2 significant figures 1 mark
Source: AQA A-Level Chemistry past papers.
What changes the value of Kc
Here is the idea that separates the strong answers from the rest: only temperature changes the value of Kc. To see why, think of Kc as the exact products-to-reactants ratio the reaction is aiming for.
Concentration, pressure and catalysts don’t change that target. They only knock the mixture away from it for a moment. Add more reactant and the ratio momentarily dips below Kc; the position shifts towards products just far enough to rebuild the same value. The system always comes back to the same number, because you never changed what it was aiming for — you only moved the position.
Temperature is different because it changes the target itself. Heating speeds up both the forward and reverse reactions, but not equally: the endothermic direction has the higher energy barrier, so it is sped up more. That tips the balance permanently and the mixture settles at a new ratio — a new Kc. For an exothermic forward reaction the reverse (endothermic) direction gains the most, so raising the temperature decreases Kc; for an endothermic forward reaction, it increases Kc.
- Only temperature changes Kc. Concentration, pressure and catalysts do not — they shift the position, and the same Kc is restored. Examiner reports flag this position-versus-constant confusion as the topic’s biggest source of lost marks.
- Match the direction to ΔH. Exothermic forward → Kc falls as temperature rises; endothermic forward → Kc rises.
🧪 Exam-style questions
For the industrial production of ethanol from ethene at 300 °C: C2H4(g) + H2O(g) ⇌ C2H5OH(g), ΔH = −46 kJ mol−1. Which statement is correct? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Kp is examined at full A-Level only (Year 13). If you’re in Year 12 (AS), you already have everything you need above this line — Kp builds on exactly the same ideas, using partial pressures in place of concentrations.
Partial pressures and Kp A-level only
For a gas-phase equilibrium, A-level swaps concentrations for partial pressures and writes the constant as Kp. Every idea from Kc carries straight over; only the currency changes.
Partial pressure of a gas = its mole fraction × the total pressure, where the mole fraction of a gas is its moles ÷ the total moles of gas. Kp = partial pressures of the products over the reactants, each raised to its balancing number.
The mole fractions of all the gases add up to 1, so the partial pressures add up to the total pressure. The routine mirrors Kc: amounts → mole fractions → partial pressures (× total pressure) → substitute → derive units.
An equilibrium mixture at a total pressure of 200 kPa contains a gas with mole fraction 0.25. Find its partial pressure.
partial pressure = mole fraction × total pressure = 0.25 × 200 = 50 kPa
For 2SO2(g) + O2(g) ⇌ 2SO3(g), an equilibrium mixture has partial pressures p(SO2) = 1.67 × 102 kPa, p(O2) = 1.02 × 102 kPa, p(SO3) = 1.85 × 102 kPa.
Step 1 — write the expression in partial pressures:
Kp = p(SO3)2 ÷ [p(SO2)2 × p(O2)]
Step 2 — substitute and evaluate:
Kp = (185)2 ÷ [(167)2 × 102] = 1.20 × 10−2 kPa−1
Units from the expression: kPa2 ÷ (kPa2 × kPa) = kPa−1. No square brackets anywhere — these are pressures.
- Partial pressures, never square brackets. Examiner reports flag square brackets in Kp — brackets mean concentration and belong to Kc. Write p(X) or pX.
- Mole fractions of the gases sum to 1. Use only the gaseous species; the partial pressures then sum to the total pressure.
- Kp has no units when the moles of gas balance — derive the units from the expression each time rather than assuming.
Interactive — work Kp one step at a time
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Step 1 — mole fraction of each gas (its moles ÷ the total moles of gas):
-
Step 2 — partial pressure of each gas (mole fraction × total pressure), in kPa:
-
Step 3 — substitute the partial pressures into Kp (never square brackets):
-
Step 4 — the units of Kp (cancel the pressure units in the expression):
🧪 Exam-style questions
For 2SO2(g) + O2(g) ⇌ 2SO3(g), 2.00 mol of SO2 is mixed with 2.00 mol of O2. The total amount of the three gases at equilibrium is 3.40 mol. What is the mole fraction of sulfur trioxide? Tick (✓) one box.
State why the equilibrium constant Kp for H2O(g) + CO(g) ⇌ H2(g) + CO2(g) has no units.
Show answer
There are equal moles of gas on both sides (2 and 2), so the pressure units in the Kp expression cancel out. 1 mark
Source: AQA A-Level Chemistry past papers.
What changes Kp A-level only
Kp obeys exactly the same rule as Kc: only temperature changes its value. This is where the topic’s biggest misconception lives, so it is worth stating flatly.
Change the pressure and the position of equilibrium shifts (to the side with fewer moles of gas) — but the individual partial pressures rearrange so that Kp comes back to the same value. A catalyst changes neither. Only temperature gives Kp a new value: for an endothermic forward reaction Kp rises with temperature, so comparing Kp at two temperatures lets you deduce the sign of ΔH.
- Changing pressure changes the position, but NOT Kp — only temperature does. Examiner reports single this out as the most common equilibrium error.
- A catalyst changes the rate of attainment, not Kp. It gets you to the same equilibrium faster.
- Deducing ΔH: a larger Kp at a higher temperature means the forward reaction is endothermic (and the converse).
🧪 Exam-style questions
Nitrogen reacts with hydrogen in this exothermic reaction: N2(g) + 3H2(g) ⇌ 2NH3(g). Which change increases the equilibrium yield of ammonia but has no effect on the value of Kp? Tick (✓) one box.
For 2SO2(g) + O2(g) ⇌ 2SO3(g), what is the effect on the value of Kp if the pressure of the equilibrium mixture is increased at constant temperature?
Show answer
Kp stays the same. 1 mark
Pressure shifts the position (here, towards SO3), but only temperature changes the value of Kp.
State the effect, if any, of adding a catalyst on the value of Kp for a gaseous equilibrium. Explain your answer.
Show answer
Effect: no effect. 1 mark
A catalyst increases the rate of the forward and reverse reactions equally, so it does not change the position of equilibrium (only temperature affects Kp). 1 mark
For 2SO3(g) ⇌ 2SO2(g) + O2(g), Kp = 7.62 × 105 Pa at 1050 K and 3.94 × 104 Pa at 500 K. Explain how this information shows that the forward reaction is endothermic.
Show answer
Kp is larger at the higher temperature. 1 mark
A higher Kp means the position has shifted further towards the products as temperature rose, which happens when the forward reaction is endothermic. 1 mark
Source: AQA A-Level Chemistry past papers.
Capstone: position vs constant A-level only
The whole topic reduces to one grid. For every change you can make to a gaseous equilibrium, ask two separate questions: does it move the position, and does it change the value of the constant? They are not the same question — and that is exactly what the marks reward.
| Change (exothermic forward reaction) | Position of equilibrium | Value of Kc | Value of Kp |
|---|---|---|---|
| Add more reactant | shifts towards products | unchanged | unchanged |
| Add more product | shifts towards reactants | unchanged | unchanged |
| Increase total pressure (gases) | shifts to the side with fewer moles of gas | unchanged | unchanged |
| Increase the temperature | shifts in the endothermic (reverse) direction | decreases | decreases |
| Add a catalyst | no shift | unchanged | unchanged |
Position levers vs the one constant lever. Concentration, pressure and catalysts are position levers — they move where the balance sits, but the constant snaps back to the same value. Temperature is the only lever that changes the constant itself, and it changes the position too.
The example that catches everyone. Increase the pressure on 2SO2 + O2 ⇌ 2SO3 and the composition genuinely changes — more SO3 forms — yet Kp is exactly the same number as before. A change in composition is not a change in the constant. Keep those two ideas apart and the equilibrium marks are yours.
Interactive — does it move the position, or change the constant?
For the exothermic gaseous equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g) (3 mol of gas on the left, 2 mol on the right), decide what each change does — make one pick in each row.
effect on the position
effect on the value of Kc / Kp
- Dynamic equilibrium (closed system): forward and reverse rates are equal; concentrations stay constant — constant, not equal.
- Le Chatelier’s principle: a change to a system at equilibrium shifts the position of equilibrium to oppose it. Answer in two steps — direction of shift, then the consequence.
- Concentration: add a species → shifts away from it. Pressure ↑ → shifts to the side with fewer moles of gas. Temperature ↑ → shifts in the endothermic direction. Catalyst: no effect on position or yield.
- Compromise conditions: real processes trade yield against rate and cost — a moderate temperature (economic rate, acceptable yield) and pressure, with a catalyst to allow a lower temperature.
- Kc = products over reactants, each concentration [X] in mol dm−3 raised to its balancing number; convert moles to concentration (÷ volume) before substituting, and derive the units each time.
- Only temperature changes the value of Kc (or Kp). Concentration, pressure and catalysts change the position, then the same K is re-established. Exothermic forward → K falls as temperature rises.
- Kp (A-level): partial pressure = mole fraction × total pressure; Kp uses partial pressures, never square brackets; gas phase only; derive units each time (none when moles of gas balance).
- Pressure changes the position but NOT Kp — only temperature does. A catalyst changes the rate of attainment, not the value of the constant.